Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-6x+y &= -4 \\ 8x+3y &= 1\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $3y = -8x+1$ Divide both sides by $3$ to isolate $y$ $y = {-\dfrac{8}{3}x + \dfrac{1}{3}}$ Substitute this expression for $y$ in the first equation. $-6x+({-\dfrac{8}{3}x + \dfrac{1}{3}}) = -4$ $-6x - \dfrac{8}{3}x + \dfrac{1}{3} = -4$ Simplify by combining terms, then solve for $x$ $-\dfrac{26}{3}x + \dfrac{1}{3} = -4$ $-\dfrac{26}{3}x = -\dfrac{13}{3}$ $x = \dfrac{1}{2}$ Substitute $\dfrac{1}{2}$ for $x$ back into the top equation. $-6( \dfrac{1}{2})+y = -4$ $-3+y = -4$ $y = -1$ $y = -1$ The solution is $\enspace x = \dfrac{1}{2}, \enspace y = -1$.